Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 115 / 1

The tangent bundle is $TM={⨆}_{p\in M}{T}_{p}M$, with its smooth vector-bundle structure from coordinate charts. A vector field is a smooth section of this bundle. The Lie bracket of vector fields is their commutator as derivations:In coordinates, $[X,Y{]}^k=X^j{\partial }_j{Y}^k-Y^j{\partial }_j{X}^k$.

The four numbered conditions here are hypotheses of one argument, so the following sections assemble its proof. Let $j:\partial M\hookrightarrow M$. Because all $X_i$ are tangent to the boundary, evaluating $\eta ={\iota }_{X_1}\cdots {\iota }_{X_n}\omega$ on a tangent vector $Y$ there evaluates the volume form on $n+1$ vectors in the $n$-dimensional boundary tangent space. Alternation gives $\boxed{j^{\ast }\eta =0}$. This uses the interior product of a differential form and will make the boundary restriction of the de Rham cohomology class vanish.

At any point, the linearly independent vectors $X_1,\dots ,X_n$ can be completed by $Y$ to a basis of $T_{p}M$. A nowhere-zero volume form is nonzero on every basis, soThus $\boxed{\eta \text{ is nowhere zero}}$. In particular, if it is an exact differential form $df$, its potential has no critical points. This is the role of pointwise independence in the boundary obstruction for commuting volume-preserving vector fields.

Let ${\Phi }_t,{\Psi }_s$ be the local flows of $X,Y$. The derivative of the pullback of $Y$ by ${\Phi }_t$ is ${\Phi }_t^{\ast }[X,Y]$. Therefore $[X,Y]=0$ makes $Y$ invariant under the $X$ flow. For fixed $t$, the two curves ${\Phi }_t({\Psi }_s(p))$ and ${\Psi }_s({\Phi }_t(p))$ then solve the same initial-value problem for $Y$. Uniqueness gives commuting flows wherever both compositions are defined. Conversely, commutation implies invariance of $Y$ under ${\Phi }_t$, and differentiation at $t=0$ gives $\boxed{[X,Y]=0}$. This proves vanishing Lie bracket is equivalent to commuting local flows.

For the contraction $\eta$, commute each Lie derivative past the other interior products using $[{\mathcal{L}}_{X_i},{\iota }_{X_j}]={\iota }_{[X_i,X_j]}=0$. Each ${\mathcal{L}}_{X_i}\omega$ is zero by volume preservation. Cartan's magic formula $d{\iota }_X={\mathcal{L}}_X-{\iota }_{X}d$, applied successively, now givesThe last term vanishes since $\omega$ has top degree. Hence $\boxed{\eta \text{ is closed}}$.

By the preceding sections, $\eta$ is a nowhere-zero closed differential form and $j^{\ast }\eta =0$. Injectivity of $j^{\ast }:H_{\mathrm{dR}}^1(M)\to H_{\mathrm{dR}}^1(\partial M)$ gives $[\eta ]=0$. The definition of de Rham cohomology then givesfor a smooth $f$, with no critical points. If the boundary were connected, $j^{\ast }df=0$ would make $f$ constant there. A nonconstant function on a compact manifold attains both extrema; a value differing from the common boundary value gives an interior extremum, contradicting $df\ne 0$. If $f$ were constant everywhere there would be the same contradiction. Empty boundary is also impossible, because an extremum would necessarily be interior. Thus $\boxed{\partial M\text{ has at least two connected components}}$, the boundary obstruction for commuting volume-preserving vector fields.

The solid torus $S^1\times D^2$ retracts onto its first circle. Its degree-one de Rham cohomology is generated by that circle's angular one-form, which restricts to a nonzero class on $S^1\times S^1$; hence the restriction map is injective. Its boundary is connected. Extensions of the two specified coordinate vector fields would be tangent there and would satisfy all the forbidden conditions. Therefore no such pair of extensions exists, for any volume form.